# Conditional expectation conditional on exponential random variable

Given $X_1$, $X_2$ are independent EXP(1) random variables, how do I compute the following expectations:

• $E[X_1 | X_1 + X_2]$

• $P(X_1 > 3 | X_1 + X_2)$

• $E[X_1 | min(X_1, t)]$

• Is this a homework question? If so, it should be flagged as such. – Placidia Dec 9 '12 at 1:35

I'll do the first one. Please, try to find similar solutions to the others. We know that $X_1$ and $X_2$ are IID (their specific distributions don't matter yet). Hence, by symmetry we have $$\textrm{E}[X_1\mid X_1+X_2] = \textrm{E}[X_2\mid X_1+X_2] \quad \textrm{a.s.}$$ Therefore, using the properties of the conditional expectation, we find $$\textrm{E}[X_1\mid X_1 + X_2] = \frac{1}{2} \Bigg( \textrm{E}[X_1\mid X_1+X_2] + \textrm{E}[X_2\mid X_1+X_2] \Bigg)$$ $$= \frac{1}{2} \textrm{E}[X_1 + X_2 \mid X_1+X_2] =\frac{X_1+X_2}{2} \quad \textrm{a.s.}$$ Since $X_1$ and $X_2$ are independent $\textrm{Exp}(1)\sim\textrm{Gamma}(1,1)$, we know that $$X_1+X_2\sim\textrm{Gamma}(2,1) \, ,$$ yielding $$\textrm{E}[X_1\mid X_1+X_2] \sim \textrm{Gamma}(2,1/2) \, .$$ It's always good to make a quick verification. By the tower property, we know that it must be the case that $$\textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=\textrm{E}[X_1] \, ,$$ but from what we've just found, we have $\textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=1$, and also $\textrm{E}[X_1]=1$.
• May I know to calculate $E[X_1 | min(X_1, t)]$, do we have to go through the standard procedure, that is, find the joint distribution of $(X_1 , min(X_1, t))$ and do integration? – John Feb 15 '16 at 15:13